Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{t^2 + t - 90}{3t - 18} \div \dfrac{-8t - 80}{t - 6} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{t^2 + t - 90}{3t - 18} \times \dfrac{t - 6}{-8t - 80} $ First factor the quadratic. $q = \dfrac{(t + 10)(t - 9)}{3t - 18} \times \dfrac{t - 6}{-8t - 80} $ Then factor out any other terms. $q = \dfrac{(t + 10)(t - 9)}{3(t - 6)} \times \dfrac{t - 6}{-8(t + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (t + 10)(t - 9) \times (t - 6) } { 3(t - 6) \times -8(t + 10) } $ $q = \dfrac{ (t + 10)(t - 9)(t - 6)}{ -24(t - 6)(t + 10)} $ Notice that $(t - 6)$ and $(t + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(t + 10)}(t - 9)(t - 6)}{ -24(t - 6)\cancel{(t + 10)}} $ We are dividing by $t + 10$ , so $t + 10 \neq 0$ Therefore, $t \neq -10$ $q = \dfrac{ \cancel{(t + 10)}(t - 9)\cancel{(t - 6)}}{ -24\cancel{(t - 6)}\cancel{(t + 10)}} $ We are dividing by $t - 6$ , so $t - 6 \neq 0$ Therefore, $t \neq 6$ $q = \dfrac{t - 9}{-24} $ $q = \dfrac{-(t - 9)}{24} ; \space t \neq -10 ; \space t \neq 6 $